If ∫01 ( e t dt /1+ t )= A then the value of ∫01 ( e t dt /(1+ t )2) is :

Question

If ∫01 ( e t dt /1+ t )= A then the value of ∫01 ( e t dt /(1+ t )2) is : (A) A +( e /2)-1  (B)  A -( e /2)+1 (C) A -( e /2)-1 (D) A +( e /2)+1

Tardigrade Admin · Accepted Answer

I=∫ limits01 (et(t+1-t)/(1+t)2) d t=∫ limits01 (et/1+t) d t-∫ limits01 et((1/1+t)-(1/(1+t)2)) d t .= A -( e t /1+ t )]01= A -( e /2)+1; Alternatively I. B. P. directly.

Q. If $\int_{0}^{1} \frac{e ^{t} d t}{1 + t} = A$ then the value of $\int_{0}^{1} \frac{e ^{t} d t}{( 1 + t ) ^{2}}$ is :

$A + \frac{e}{2} - 1$

$A - \frac{e}{2} + 1$

$A - \frac{e}{2} - 1$

$A + \frac{e}{2} + 1$

$I = 0 \int 1 \frac{e ^{t} ( t + 1 - t )}{( 1 + t ) ^{2}} d t = 0 \int 1 \frac{e ^{t}}{1 + t} d t - 0 \int 1 e^{t} (\frac{1}{1 + t} - \frac{1}{( 1 + t ) ^{2}}) d t$
$= A - \frac{e ^{t}}{1 + t}]_{0}^{1} = A - \frac{e}{2} + 1$ ; Alternatively I. B. P. directly.

Q. If ∫01​1+tetdt​=A then the value of ∫01​(1+t)2etdt​ is :

A+2e​−1

A−2e​+1

A−2e​−1

A+2e​+1